Answer :
Final answer:
To estimate the freezing point of the ethylene glycol solution, calculate the molality by converting grams to moles of solute and kilogram of solvent, and then use the freezing point depression formula with the freezing point depression constant for water. The calculated freezing point of the ethylene glycol solution would be -6.564°C.
Explanation:
To determine the freezing point of an ethylene glycol solution containing 21.4 g of ethylene glycol (C₂H₆O₂) in 97.6 mL of water, one must first calculate the molality of the solution. Molality is defined as the number of moles of solute per kilogram of solvent.
First, calculate the moles of ethylene glycol using its molar mass (62.07 g/mol):
21.4 g C₂H₆O₂ / 62.07 g/mol = 0.3445 mol C₂H₆O₂
Assuming the density of water is 1.00 g/mL, 97.6 mL of water is equivalent to 97.6 g. Since we need the mass in kilograms for molality:
97.6 g water * (1 kg / 1000 g) = 0.0976 kg water
The molality (m) is thus:
0.3445 mol C₂H₆O₂ / 0.0976 kg water = 3.529 mol/kg
The freezing point depression (ΔTf) can then be calculated using the formula:
ΔTf = i * Kf * m
Where i is the van't Hoff factor (which is 1 for ethylene glycol because it does not dissociate in solution), Kf is the freezing point depression constant for water (1.86°C kg/mol), and m is the molality calculated above.
Applying the values:
ΔTf = 1 * 1.86°C kg/mol * 3.529 mol/kg = 6.564°C
The freezing point of the solution is thus the freezing point of pure water (0°C) minus the freezing point depression:
0°C - 6.564°C = -6.564°C
This would be the estimate for the freezing point of the ethylene glycol solution in degrees Celsius.